If  $af(\tan x) + bf (cot x) = x$ then $f\text{'}(cot x) =$

af(tanx) + bf(cotx) = x  (1)
Replace x by $\frac{\mathrm{\pi }}{2}-\mathrm{x}$
$\begin{array}{l}\mathrm{af}\left(\tan \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)\right)+\mathrm{bf}\left(\cot \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)\right)=\frac{\mathrm{\pi }}{2}-\mathrm{x}\\ \mathrm{af}(\cot \mathrm{x})+\mathrm{bf}(\tan \mathrm{x})=\frac{\mathrm{\pi }}{2}-\mathrm{x} ....\left(2\right)\end{array}$
Equation (1) Multiply with b
abf(tanx) + b²f(cotx) = bx ....... (3)
Equation (2) Multiply with a
${\mathrm{a}}^2\mathrm{f}(\cot \mathrm{x})+\mathrm{abf}(\tan \mathrm{x})=\frac{\mathrm{a\pi }}{2}-\mathrm{ax} ...\left(4\right)$
Form (3) & (4) eliminate f(tanx)
$\begin{array}{l}\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)\mathrm{f}(\cot \mathrm{x})=\frac{\mathrm{a\pi }}{2}-\mathrm{ax}-\mathrm{bx}\\ \mathrm{f}(\cot \mathrm{x})=\frac{\mathrm{a\pi }/2}{{\mathrm{a}}^2-{\mathrm{b}}^2}-\frac{(\mathrm{a}+\mathrm{b})\mathrm{x}}{(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})}\\ \mathrm{f}(\cot \mathrm{x})=\frac{\mathrm{a\pi }}{2\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}-\frac{\mathrm{x}}{\mathrm{a}-\mathrm{b}}\end{array}$
Diff on both sides
$\begin{array}{l}{\mathrm{f}}^{\mathrm{\prime }}(\cot \mathrm{x})\left(-{\mathrm{cosec}}^2\mathrm{s}\right)=\frac{-1}{\mathrm{a}-\mathrm{b}}\\ {\mathrm{f}}^{\mathrm{\prime }}(\cot \mathrm{x})=\frac{{\sin }^2\mathrm{x}}{\mathrm{a}-\mathrm{b}}\end{array}$